Real NumbersClass 10 Mathematics NCERT Solutions

(i) 140 We use the prime factorisation method: $140 = 2 \times 70$ $= 2 \times 2 \times 35$ $= 2 \times 2 \times 5 \times 7$

Final Answer: The prime factorisation of 140 is $2^2 \times 5 \times 7$.

(ii) 156 We use the prime factorisation method: $156 = 2 \times 78$ $= 2 \times 2 \times 39$ $= 2 \times 2 \times 3 \times 13$

Final Answer: The prime factorisation of 156 is $2^2 \times 3 \times 13$.

(iii) 3825 We use the prime factorisation method: $3825 = 3 \times 1275$ $= 3 \times 3 \times 425$ $= 3 \times 3 \times 5 \times 85$ $= 3 \times 3 \times 5 \times 5 \times 17$

Final Answer: The prime factorisation of 3825 is $3^2 \times 5^2 \times 17$.

(iv) 5005 We use the prime factorisation method: $5005 = 5 \times 1001$ $= 5 \times 7 \times 143$ $= 5 \times 7 \times 11 \times 13$

Final Answer: The prime factorisation of 5005 is $5 \times 7 \times 11 \times 13$.

(v) 7429 We use the prime factorisation method: $7429 = 17 \times 437$ $= 17 \times 19 \times 23$

Final Answer: The prime factorisation of 7429 is $17 \times 19 \times 23$.

Given: Integers are 26 and 91.

To Find: HCF and LCM, and verify the relationship.

Solution: First, find the prime factorisation of each number: $26 = 2 \times 13$ $91 = 7 \times 13$

HCF is the product of the smallest power of each common prime factor. $\text{HCF}(26, 91) = 13$

LCM is the product of the greatest power of each prime factor involved in the numbers. $\text{LCM}(26, 91) = 2 \times 7 \times 13 = 182$

Verification: $\text{HCF} \times \text{LCM} = 13 \times 182 = 2366$ Product of the two numbers $= 26 \times 91 = 2366$

Since LHS = RHS, hence verified.

Given: Integers are 510 and 92.

To Find: HCF and LCM, and verify the relationship.

Solution: Prime factorisation: $510 = 2 \times 255 = 2 \times 3 \times 85 = 2 \times 3 \times 5 \times 17$ $92 = 2 \times 46 = 2 \times 2 \times 23 = 2^2 \times 23$

$\text{HCF}(510, 92) = 2$ $\text{LCM}(510, 92) = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$

Verification: $\text{HCF} \times \text{LCM} = 2 \times 23460 = 46920$ Product of the two numbers $= 510 \times 92 = 46920$

Since LHS = RHS, hence verified.

Given: Integers are 336 and 54.

To Find: HCF and LCM, and verify the relationship.

Solution: Prime factorisation: $336 = 2 \times 168 = 2^2 \times 84 = 2^3 \times 42 = 2^4 \times 21 = 2^4 \times 3 \times 7$ $54 = 2 \times 27 = 2 \times 3^3$

$\text{HCF}(336, 54) = 2 \times 3 = 6$ $\text{LCM}(336, 54) = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024$

Verification: $\text{HCF} \times \text{LCM} = 6 \times 3024 = 18144$ Product of the two numbers $= 336 \times 54 = 18144$

Since LHS = RHS, hence verified.

Given: Integers are 12, 15, and 21.

To Find: HCF and LCM.

Solution: Prime factorisation: $12 = 2 \times 6 = 2^2 \times 3$ $15 = 3 \times 5$ $21 = 3 \times 7$

HCF is the product of the smallest power of each common prime factor. The only common prime factor is 3. $\text{HCF}(12, 15, 21) = 3$

LCM is the product of the greatest power of each prime factor involved. $\text{LCM}(12, 15, 21) = 2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420$

Final Answer: HCF = 3, LCM = 420.

Given: Integers are 17, 23, and 29.

To Find: HCF and LCM.

Solution: Prime factorisation: $17 = 1 \times 17$ $23 = 1 \times 23$ $29 = 1 \times 29$ All three numbers are prime numbers. Their only common factor is 1. $\text{HCF}(17, 23, 29) = 1$

The LCM is the product of the numbers themselves. $\text{LCM}(17, 23, 29) = 17 \times 23 \times 29 = 11339$

Final Answer: HCF = 1, LCM = 11339.

Given: Integers are 8, 9, and 25.

To Find: HCF and LCM.

Solution: Prime factorisation: $8 = 2 \times 2 \times 2 = 2^3$ $9 = 3 \times 3 = 3^2$ $25 = 5 \times 5 = 5^2$ There are no common prime factors among the three numbers. $\text{HCF}(8, 9, 25) = 1$

The LCM is the product of the numbers as they are coprime. $\text{LCM}(8, 9, 25) = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800$

Final Answer: HCF = 1, LCM = 1800.

Given: The two numbers are 306 and 657. $\text{HCF}(306, 657) = 9$

To Find: $\text{LCM}(306, 657)$

Formula: For any two positive integers a and b, we have: $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$

Solution: Using the formula, we can write: $\text{LCM}(306, 657) = \frac{306 \times 657}{\text{HCF}(306, 657)}$

Substituting the given value of HCF: $\text{LCM}(306, 657) = \frac{306 \times 657}{9}$

$= 34 \times 657$ $= 22338$

Final Answer: The LCM of 306 and 657 is 22338.

To Check: Whether $6^n$ can end with the digit 0 for any natural number $n$.

Reasoning: A number ends with the digit 0 if it is divisible by 10. This means its prime factorisation must include both 2 and 5.

Solution: Let us find the prime factors of $6^n$. $6^n = (2 \times 3)^n = 2^n \times 3^n$

The prime factors of $6^n$ are only 2 and 3. For any natural number $n$, the prime factorisation of $6^n$ will only contain the primes 2 and 3.

The prime factor 5 is not present in the prime factorisation of $6^n$. According to the Fundamental Theorem of Arithmetic, the prime factorisation of any number is unique. Therefore, $6^n$ can never have 5 as a factor.

Since $6^n$ is not divisible by 5, it cannot end with the digit 0.

Final Answer: There is no natural number $n$ for which $6^n$ ends with the digit zero.

To Explain: Why the given expressions represent composite numbers.

Definition: A composite number is a positive integer that has at least one divisor other than 1 and itself.

First number: $7 \times 11 \times 13 + 13$ We can take 13 as a common factor from the expression: $7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1)$ $= 13 \times (77 + 1)$ $= 13 \times 78$

The given expression can be written as a product of two factors, 13 and 78. Since the number has factors other than 1 and itself, it is a composite number.

Second number: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ We can take 5 as a common factor from the expression: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)$ $= 5 \times (1008 + 1)$ $= 5 \times 1009$

The given expression can be written as a product of two factors, 5 and 1009. Since the number has factors other than 1 and itself, it is a composite number.

Conclusion: Both numbers are composite because they can be expressed as a product of factors other than 1 and the number itself.

Given: Time taken by Sonia for one round = 18 minutes. Time taken by Ravi for one round = 12 minutes. They start at the same point, at the same time, and go in the same direction.

To Find: The time after which they will meet again at the starting point.

Solution: To find the time when they will meet again at the starting point, we need to find the Least Common Multiple (LCM) of the time taken by each of them. The LCM of 18 and 12 will give the earliest time at which both will be at the starting point together.

First, find the prime factorisation of 18 and 12: $18 = 2 \times 9 = 2 \times 3^2$ $12 = 2 \times 6 = 2^2 \times 3$

Now, find the LCM by taking the highest power of each prime factor present in the numbers. $\text{LCM}(18, 12) = 2^2 \times 3^2$ $= 4 \times 9$ $= 36$

This means that after 36 minutes, Sonia will have completed $36/18 = 2$ rounds, and Ravi will have completed $36/12 = 3$ rounds. They will both be at the starting point.

Final Answer: They will meet again at the starting point after 36 minutes.

To Prove: $\sqrt{5}$ is an irrational number.

Proof: We will use the method of proof by contradiction. Let us assume, to the contrary, that $\sqrt{5}$ is a rational number.

If it is rational, it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and $a$ and $b$ are coprime (they have no common factors other than 1). So, $\sqrt{5} = \frac{a}{b}$.

Squaring both sides, we get: $5 = \frac{a^2}{b^2}$ $5b^2 = a^2 \quad ...(1)$

This implies that $a^2$ is divisible by 5. By the theorem that if a prime $p$ divides $a^2$, then $p$ divides $a$, it follows that $a$ is also divisible by 5. So, we can write $a = 5c$ for some integer $c$.

Substitute $a = 5c$ in equation (1): $5b^2 = (5c)^2$ $5b^2 = 25c^2$ $b^2 = 5c^2$

This implies that $b^2$ is divisible by 5. Therefore, $b$ is also divisible by 5.

So, both $a$ and $b$ are divisible by 5. This means that $a$ and $b$ have at least 5 as a common factor. But this contradicts our initial assumption that $a$ and $b$ are coprime.

This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational. Therefore, our assumption is wrong.

Hence Proved: $\sqrt{5}$ is an irrational number.

To Prove: $3 + 2\sqrt{5}$ is an irrational number.

Proof: We will use the method of proof by contradiction. Let us assume, to the contrary, that $3 + 2\sqrt{5}$ is a rational number.

If it is rational, it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers and $b \neq 0$. $3 + 2\sqrt{5} = \frac{a}{b}$

Now, we rearrange the equation to isolate $\sqrt{5}$: $2\sqrt{5} = \frac{a}{b} - 3$ $2\sqrt{5} = \frac{a - 3b}{b}$ $\sqrt{5} = \frac{a - 3b}{2b}$

Since $a$ and $b$ are integers, $a - 3b$ is an integer and $2b$ is a non-zero integer. Therefore, $\frac{a - 3b}{2b}$ is a rational number.

This implies that $\sqrt{5}$ is a rational number. However, we know that $\sqrt{5}$ is an irrational number. This is a contradiction.

This contradiction has arisen because of our incorrect assumption that $3 + 2\sqrt{5}$ is rational. Therefore, our assumption is wrong.

Hence Proved: $3 + 2\sqrt{5}$ is an irrational number.

Proof: Let us assume, to the contrary, that $\frac{1}{\sqrt{2}}$ is rational. Then, we can find coprime integers $a$ and $b$ ($b \neq 0$) such that: $\frac{1}{\sqrt{2}} = \frac{a}{b}$ Rearranging the equation, we get: $\sqrt{2} = \frac{b}{a}$ Since $a$ and $b$ are integers, $\frac{b}{a}$ is a rational number. This implies that $\sqrt{2}$ is rational. But this contradicts the fact that $\sqrt{2}$ is irrational. This contradiction has arisen because of our incorrect assumption.

Hence Proved: $\frac{1}{\sqrt{2}}$ is irrational.

Proof: Let us assume, to the contrary, that $7\sqrt{5}$ is rational. Then, we can find coprime integers $a$ and $b$ ($b \neq 0$) such that: $7\sqrt{5} = \frac{a}{b}$ Rearranging the equation, we get: $\sqrt{5} = \frac{a}{7b}$ Since $a$, $b$, and 7 are integers, $\frac{a}{7b}$ is a rational number. This implies that $\sqrt{5}$ is rational. But this contradicts the fact that $\sqrt{5}$ is irrational. This contradiction has arisen because of our incorrect assumption.

Hence Proved: $7\sqrt{5}$ is irrational.

Proof: Let us assume, to the contrary, that $6+\sqrt{2}$ is rational. Then, we can find coprime integers $a$ and $b$ ($b \neq 0$) such that: $6+\sqrt{2} = \frac{a}{b}$ Rearranging the equation, we get: $\sqrt{2} = \frac{a}{b} - 6$ $\sqrt{2} = \frac{a - 6b}{b}$ Since $a$ and $b$ are integers, $\frac{a - 6b}{b}$ is a rational number. This implies that $\sqrt{2}$ is rational. But this contradicts the fact that $\sqrt{2}$ is irrational. This contradiction has arisen because of our incorrect assumption.

Hence Proved: $6+\sqrt{2}$ is irrational.